4一阶线性常系数微分方程组 dxi(t a1()+a12x2(D)+…+a1nxn(t)+f1(1) dt dx2(t dx=的2ux1(t)+a2(+…+a2nxn()+/2( dxn(t) anIx(t)+an2x2(t)+.+annen(t)+fn(t) dt 满足初始条件x(0)=C,i=1,2,…,n
4 一阶线性常系数微分方程组 1 11 1 12 2 1 1 2 21 1 22 2 2 2 1 1 2 2 d ( ) ( ) ( ) ( ) ( ) d d ( ) ( ) ( ) ( ) ( ) d d ( ) ( ) ( ) ( ) ( ) d n n n n n n n nn n n x t a x t a x t a x t f t t x t a x t a x t a x t f t t x t a x t a x t a x t f t t = + + + + = + + + + = + + + + 0 ( ) , 1,2, , 满足初始条件 x t c i n i i = =
dx(t =Ax(t)+∫(t) →dt x(to)=C 其中,A=Vxm,x()=(x(t),x2(),…,x( c=(a1,c2,…,cn)1,f()=(f(),2(t)…,fn() →(ex(1)=e(-4)x(+4dx(t) dr =e Atr delt d4x(t)=eAf)在,1上积分
0 d ( ) ( ) ( ) d ( ) x t Ax t f t t x t c = + = ( ) , 其中,A a = ij n n 1 2 ( ) ( ( ), ( ), , ( )) , T n x t x t x t x t = 1 2 ( , , , ) , T n c c c c = 1 2 ( ) ( ( ), ( ), , ( )) , T n f t f t f t f t = d d ( ) ( ( )) ( ) ( ) d d At At At x t e x t e A x t e t t − − − = − + d ( ) ( ( )) ( ) d At At x t e Ax t e f t t − − = − = 0 在 上积分 [ , ] t t
e-Alx(0-e Aox(t0)=eAt f(cjd →x0=+c+ur 例1:求解初值问题 dx(t) x1(t)-2x2(t)+6x3(t)- dt dx2(t) =-x1()+3x3() dt dx3() dt x1()-x2()+4x3()+e 1(0)=1,x2(0)=0,x3(0)=0
1 1 2 3 2 1 3 3 1 2 3 1 2 3 d ( ) ( ) 2 ( ) 6 ( ) d d ( ) ( ) 3 ( ) d d ( ) ( ) ( ) 4 ( ) d (0) 1, (0) 0, (0) 0 t t x t x t x t x t e t x t x t x t t x t x t x t x t e t x x x = − − + − = − + = − − + + = = = 例 1: 求解初值问题0 0 0 ( ) ( ) ( )d At t At A t e x t e x t e f − − − − = 0 0 ( ) ( ) ( )d A t t t At A t x t e c e e f − − = +
解 1-26 A=-103,c=0,f(r)=0|→ 1-2t-2t6t At =e 3t, t1+3t 1-2t 1-87 e c=e-t en f( dE 4τdz→ 0
解: 1 2 6 1 1 0 3 , 0 , ( ) 0 1 1 4 0 t t e A c f t e − − − = − = = − − 1 2 2 6 1 3 , 1 3 At t t t t e e t t t t t t − − = − − − − + 1 2 At t t e c e t t − = − − 0 ( )d t A e f − 0 1 8 4 d 1 4 t − − = − −
7 t-2t 2t2+2 x(t=e C+eat rt e f∫(z)dτ 1-3t+4t x(t=ec+ At e f(rdr=e-t+2r 2
0 ( )d t A e f − 2 2 2 4 2 2 t t t t t − − = − − 0 ( )d t At A e e f − 2 2 2 4 1 2 2 2 t t e t t − = + 0 0 ( ) ( ) ( )d A t t t At A t x t e c e e f − − = + 0 ( ) ( )d t At At A x t e c e e f − = + 2 2 2 1 3 4 2 2 t t t e t t t − + = − +
定义设堤是阶常系数矩阵,如果对任意的0 和x0,初值问题 dx(t) Ax(t) dt x(to)=xo 的解x(满足imx()=0,则称d(盐=Ax()的 →+0 dt 解是渐进稳定的
0 定义 设 是 阶常系数矩阵 如果对任意的 A n t , 和 ,初值问题 x0 0 0 d ( ) ( ) d ( ) x t Ax t t x t x = = d ( ) ( ) lim ( ) 0 ( ) t d x t x t x t Ax t →+ t 的解 满足 ,则称 的 = = 解是渐进稳定的
定理对任意的0和x,初值问题 dx(t)=Ax(t),x(to)=xo dt 的解κ(t)是渐进稳定的充要条件是的特征值都 有负实部 证:必要性:451=4151(1=+iB1,01≥0)→ dx(t) Ax(t),x(0)=5的解为x(t)=e51 is oat(cos bit+isin Bt )51t x(t)不收敛(矛盾
0 定理 对任意的t 和 ,初值问题 x0 0 0 d ( ) ( ), ( ) d x t Ax t x t x t = = 的解x t( )是渐进稳定的充要条件是 的特征值都 A 有负实部. 证: 必要性: 1 1 1 1 1 1 1 A i = = + ( , 0) 1 d ( ) ( ), (0) d x t Ax t x t = = 的解为 1 ( ) At x t e = 1 1 t e = 1 1 1 1 cos sin ) t e t i t = + ( t → x t( )不收敛(矛盾)
充分性:对任意的和vm初值问题=Ax(O dt x(o)=x0的解为x()=4(4)104(4都有负实部 lim x(t=lim ea()xo=0 t→+
充分性: 0 对任意的t 和 ,初值问题 x0 d ( ) ( ), d x t Ax t t = x t x ( ) 0 0 = 的解为 0 ( ) 0 ( ) A t t x t e x − = ( ) A 都有负实部 0 ( ) 0 lim ( ) lim 0 A t t t t x t e x − →+ →+ = =